Conductors and minimal discriminants of
hyperelliptic curves
Padmavathi Srinivasan
MIT
AMS Summer Institute in Algebraic Geometry, Salt Lake City
July 31, 2015
Outline
1
Introduction
2
Conductors and minimal discriminants
3
Comparing conductors and discriminants
What are conductors and minimal discriminants?
Degenerating family
of hyperelliptic curves
Measures of
degeneracy
Artin conductor
Minimal discriminant
How are these related?
y 2 = (x − 1)(x 2 − t)
y2 = x3 − t
Outline
1
Introduction
2
Conductors and minimal discriminants
3
Comparing conductors and discriminants
Notation
R : complete discrete valuation ring
K : fraction field of R
k : residue field of R, algebraically closed, char 6= 2
ν : discrete valuation K → Z ∪ {∞}
t : a uniformizer of R, i.e., ν(t) = 1.
Examples: C[[t]], Zunr
p
Notation
f (x) : monic, squarefree, even degree ≥ 4 polynomial in R[x]
X : smooth projective model of the plane curve y 2 = f (x) over K
g:
genus of X
If f (x) = x d + ad−1 x d−1 + . . . + a0 factors as (x − α1 ) . . . (x − αd )
in K [x], then
Y
disc(f ) :=
(αi − αj )2
i<j
∈ K [a0 , . . . , ad−1 ]
Minimal discriminant of X
f (x) = x d + ad−1 x d−1 + . . . + a0 = (x − α1 ) . . . (x − αd )
The naive disciminant of y 2 = f (x) is defined to be
Y
∆f := ν(disc(f )) = ν
(αi − αj )2 .
i<j
The minimal discriminant ∆min of X is given by
!
f (x) ∈ R[x] such that y 2 = f (x)
∆min = ∆min (X ) := min ∆f is birational to X over K
∆min = 0 ⇐⇒ X has good reduction.
Artin conductor
X : a hyperelliptic curve over K
X : a proper, flat, regular R-scheme with XK ' X
XK : geometric generic fiber of X
Xk : special fiber of X
Fix a prime ` 6= char k. For any curve C over an algebraically
closed field of char 6= `, let
P
i (C , Q )
χ(C ) := 2i=0 (−1)i dim Hét
`
δ : Swan conductor for the representation H 1 (XK , Q` )
(integer, ≥ 0, measure of wild ramification).
Artin Conductor
X : a hyperelliptic curve over K
X : a proper, flat, regular R-scheme with XK ' X
XK : geometric generic fiber of X
Xk : special fiber of X
χ : `-adic Euler Poincaré characteristic
δ : Swan conductor for the representation H 1 (XK , Q` )
− Art(X /R) = χ(Xk ) − χ(XK ) + δ
=
(Saito)
Deligne discriminant.
Properties of the Artin Conductor
Art(X /R) is independent of `.
− Art(X /R) ≥ 0.
− Art(X /R) = 0 ⇐⇒ X is smooth or g = 1 and (Xk )red
is smooth.
Let n be the number of components of Xk and let be the
tame conductor. Then,
− Art(X /R) = (n − 1) + + δ
≥ n − 1.
When X is regular and semi-stable,
− Art(X /R) = # singular points of Xk .
Outline
1
Introduction
2
Conductors and minimal discriminants
3
Comparing conductors and discriminants
Earlier results (Ogg, Saito, Liu)
Let X be the minimal proper regular model of X .
If g = 1, then − Art(X /R) = ∆min [Ogg-Saito formula].
This also holds when char k = 2.
If g = 2, Liu showed that − Art(X /R) ≤ ∆min . He showed
that equality does not always hold.
Question: Does − Art(X /R) ≤ ∆min hold for hyperelliptic curves
of arbitrary genus?
Main result
Theorem ( )
Let X be a hyperelliptic curve over K and let X be the minimal
proper regular model of X . Assume that
(a) the Weierstrass points of X are K -rational, and,
(b) that the residue characteristic of K is not 2.
Then,
− Art(X /R) ≤ ∆min .
Explicit construction of a regular model X 0
The first step in the proof is the explicit construction of a regular
model X 0 for X (not necessarily minimal).
Lemma
Let Bl P1R be an arithmetic surface birational to P1R .
Let f be an element of the function field of P1R .
Assume that the odd multiplicity components of the divisor of f on
Bl P1R are disjoint.
p
Then, the normalization of Bl P1R in K (x, f (x)) is a proper
regular model for the hyperelliptic curve given by y 2 = f (x).
Proof strategy
1. Explicitly construct a regular model X 0 (not necessarily
minimal).
P
2. X 0 is strict simple normal crossings, Xk0 = mi Γi and
mi ∈ {1, 2}.


 X
X
X
Γi .Γj .
(1 − mi )χ(Γi ) +
(mj − 1)Γi .Γj +
− Art(X 0 /R) =


i
j6=i
3. Similarly decompose ∆min into local terms.
4. Prove a local comparison inequality.
5. Add local inequalities.
− Art(X /R) ≤ − Art(X 0 /R) ≤ ∆min .
i<j
Finally . . .
Thank you!